Negative Binomial Mean, Variance, And Dispersion

Why the Negative Binomial Has Extra Variance

A Poisson frequency model has equal mean and variance. A negative binomial model can have a larger variance than its mean, allowing claim counts to vary more widely than a same-mean Poisson count. To calculate and interpret that difference correctly, first identify exactly what the random variable counts. In particular, “failures before a fixed number of successes” and “trials until a fixed number of successes” differ by a constant, so they do not have the same mean.

The formulas below assume independent Bernoulli trials, each with the same success probability \(p\), where \(0<p<1\). For the failures-count version, the number of required successes \(r\) is a positive integer. A mean-size parameterization also permits positive noninteger \(r\).

Mean and Variance for Failures Before \(r\) Successes

One way to derive the moments is to split the waiting process into \(r\) stages. Let \(W_j\) be the number of failures after the \((j-1)\)th success and before the \(j\)th success. Each stage has the geometric distribution \(P(W_j=k)=pq^k\), and the stages are independent because the trials are independent and have the same success probability. The total failures are \(K=W_1+\cdots+W_r\).

For one stage, the geometric-series identities \(\sum_{k=0}^{\infty}q^k=1/(1-q)\), \(\sum_{k=0}^{\infty}kq^k=q/(1-q)^2\), and \(\sum_{k=0}^{\infty}k^2q^k=q(1+q)/(1-q)^3\) give its first two moments. Since \(1-q=p\),

Subtracting the square of the mean gives \(\operatorname{Var}(W_j)=q(1+q)/p^2-q^2/p^2=q/p^2\). Independence lets us add the stage means and variances. Thus,

This derivation relies on a fixed number of independent trials’ successes being sought and a constant \(p\). If the trial probabilities vary, or the stopping rule is different, these formulas do not automatically apply.

Mean-Size Parameterization and Poisson Comparison

Sometimes the model is specified by its mean \(\mu\) and size parameter \(r\), rather than by \(r\) and \(p\). Solving \(\mu=rq/p\), together with \(p+q=1\), gives \(p=r/(r+\mu)\) and \(q=\mu/(r+\mu)\). Substitution into the variance formula yields a convenient form.

The result follows directly: \(rq/p^2=r[\mu/(r+\mu)]/ [r/(r+\mu)]^2=\mu(r+\mu)/r=\mu+\mu^2/r\). A Poisson random variable with the same mean \(\mu\) has variance \(\mu\), so the difference \(\mu^2/r\) measures the negative binomial’s additional variance under this parameterization. The size \(r\) controls that difference: for fixed \(\mu\), a smaller \(r\) means a larger excess variance.

Some texts instead use a dispersion parameter \(\alpha=1/r\). In that notation, the same relationship is \(\operatorname{Var}(K)=\mu+\alpha\mu^2\). Check which parameter is being supplied before substituting: a size parameter \(r\) and a dispersion parameter \(\alpha\) are reciprocals, not interchangeable values.

When the Variable Counts Trials Instead

A second common convention counts the total number of trials \(T\) needed to obtain \(r\) successes. Every such sequence contains exactly \(r\) successes, so \(T=K+r\), where \(K\) counts failures before the \(r\)th success. Adding a constant increases the mean by that constant but leaves the variance unchanged.

Common Exam Traps

• Confusing \(p\) and \(q\). With \(p\) defined as success probability, the failures-count mean is \(rq/p\), not \(rp/q\). Define \(q=1-p\) before calculating.
• Using the trials mean for a failures count. \(r/p\) is the expected number of trials to obtain \(r\) successes. The expected number of failures is \(rq/p\).
• Assuming the variance equals the mean. That equality belongs to the Poisson model. For the failures-count negative binomial model, the variance is \(\mu+\mu^2/r\).
• Confusing size with dispersion. If the parameter is \(\alpha\) in \(\mu+\alpha\mu^2\), then \(r=1/\alpha\); substituting \(\alpha\) as though it were \(r\) reverses the effect.

The dispersion comparison is meaningful only when the two models refer to counts on the same scale and have the same mean. In particular, compare the failures count \(K\) with a Poisson count of mean \(E[K]\); do not compare \(K\) with a Poisson variable whose mean is \(E[T]\).